 Yeah. Hello. Good evening, guys. Am I audible? Hello, everyone. Yes. So last class, we discussed about this thing. Optical isomerism we were discussing, right? Yeah. So we have seen what is optical isomerism. And we were going through the terms that is involved in optical isomerism. Correct. Yes. And the terms involved in optical isomerism and the first term we had discussed is the call center. Okay. Okay, so we'll start with. We had discussed what is called center. It is a sp3 hybridized atom having four, having four different atoms or groups attached to it. Correct. Exceptions also we have seen tertiary amine is not an. Right. Next slide down the term right on chirality, chirality and chiral molecule, chirality and chiral molecule. See, first of all, do not get confused with this chirality or chiral molecule with chiral center. chiral center has nothing to do with chirality and chiral molecule. Correct. It is a different term. chiral center is a different term. Both are different terms. Okay. They have no relation. Okay. So what is a chiral molecule? A chiral molecule is the one. The molecule which cannot be divided divided to equal halves. Okay. It's called chiral molecule. Basically, there is no elements of symmetry there. No elements of symmetry. What does it mean? Means the molecule. We won't have any symmetry in the molecule. Right. Through which we cannot divide the molecule into two equal half. Right. No symmetry is there. So when the molecule has no symmetry, it is called chiral molecule. Okay. chirality is what chirality is the property of a molecule by which a molecule cannot be divided into two equal halves. Right. So chirality is what basically chirality is the property of a chiral molecule. It is the property of a chiral molecule. The molecule which is chiral has this property, chirality. What is this property? It is the property by which a molecule cannot be divided into two equal half. It's called chirality. Correct. So chirality is the property of chiral molecule. Another thing you write on chiral molecules are non-superimposable. chiral molecules are non-superimposable to their mirror image. Superimposable to their mirror image. Okay. chiral molecule. What is non-superimposable? Non-superimposable means you see first of all superimposability is what? Superimposability is a 3D phenomenon. Superimposability is the 3D phenomenon. 3D phenomenon. If two molecules are superimposable, molecules are superimposable. They look like single molecule when placed over each other. Copy this down first. Done. Now you see. Gayatri, I'll give you the definition of chiral centre weight. Okay. Pooja, I'll share. Okay. Whatever we've done last week, Pooja, you can start a flash from here. Okay. You can understand this. Okay. No problem. Gayatri, I'll give you a definition. Wait. Let me finish this. Okay. Yeah. So when placed over each other, look at this molecule. Let's try and understand what is superimposability. Okay. Okay. After this, I'll do the chiral centre weight. I got it. chiral centre definition. I'll just give you this. Let me finish this. Since you have started it, let me finish this. Okay. I'll do. Wait. So consider this molecule. Carbon. Okay. We have here, suppose CH3. And here we have OH. Here we have CL. And here we have H. You know, what is this bond means? This we call it as flying wedge representation. Okay. Flying wedge representation of a molecule. One more we have. I'll tell you that. What is that? But first you copy down this flying wedge representation. This means what this wedge that we have. It is coming out of the plane. This you must understand. It is three representation. This bond chlorine is coming out of the plane. Means you can understand the screen, the laptop screen that we have now. Chlorine is coming towards you, towards the observer. Okay. Above the screen it is. This bond is going into the plane. So it is going behind the screen away from the observer. Got it? Into the plane. This is flying wedge representation. And this two bond is present along the plane of the screen. Correct? Now. When you take the mirror image of this compound. Okay. What do you get? Suppose this is a mirror you placed side by this molecule. So what happens here? You see. This carbon will be here. Here's three on the top. And this side we have H. And then we have CL. And here we have OH. Now. If you want to know, these two compounds are superimposable or not. What do you do? You just pick this compound and place over this. What do you want to do? Pick this compound and place over this compound. So this carbon will come over this carbon. This methyl group will come over this methyl group. This chlorine will come over this chlorine. But the position of H and OH is interchanged over here. Right. So since this. Oh, which is coming over H and H is coming over OH. Hence these two compounds are non superimposable. That doesn't look like. You know. Single molecule. Hence they are not superimposable. So this molecule, the pair we have is non superimposable. This two pair are non superimposable. And since it is non superimposable to their mirror image, hence the molecule is said to be carol. Right. The molecule is said to be carol. There is no plane of symmetry. Carol molecule. Yes. Yes. Correct. If suppose you replace this OH by H. Right. Then it is superimposable. Okay. So non superimposability means this. Correct. And if this molecule you see. If you try to draw a plane of symmetry, you won't get any plane of symmetry into this. That trial molecule. There is no plane of symmetry. Whatever you want. You can try. There's no center of symmetry in this. Okay. Center of symmetry means what? You have a point in the molecule. Through which all the identical molecules will have equal distance. Which is not right. There is no plane of symmetry in this molecule. When there is no symmetry. Then the molecule is said to be carol. Okay. This is what we discussed. Hands are what? Hands are also non superimposable. You look at this. Look at this image. This is non superimposable. Okay. And then don't think of that you flip this hand and you place over it. No. We cannot flip out of the plane. Okay. Means you can not, you know, like. Simple. If you take this compound. Okay. Within the plane, the rotation is allowed. Means what? Like you, you know. You can rotate this molecule in this plane. Like this. But you cannot flip the molecule. Keep that in mind in order to understand whether the molecule is superimposable or not. This kind of rotation is allowed. You can rotate the molecule in this direction. Any direction clockwise or anticlockwise. And you can check with it coming over this molecule or not exactly. Right. But you cannot flip the molecule in this direction. You cannot flip the compound. The entire compound. Right. Same thing we have here. You see. Hands. You can. You can rotate just a second. Just a second. So this is a non superimposable. Hands are non superimposable. Mind it. You cannot flip this hand this way. Right. You cannot flip the molecule in this direction. You cannot flip the molecule in this direction. You cannot flip the molecule in this direction. You cannot flip it this way. But you can do this in this way. You cannot flip like this and place it over. Right. You have to just cross. Two hands like this. Just pick this up and place over it. Right. So this, you see, if you place this hand over this one, then this thumb. Is in this direction. Right, but this thumb will be in this direction. Hence, it is non superimposable. Correct. Always keep this mind. out of the plane rotation or flipping is not allowed. Within the plane you can rotate like the steering of the car, that you can rotate like that, okay? But flipping is not allowed. Understood? This is what superimposability means, correct? Now, when you look at this term, chiral center, chiral molecule here, so chirality is the necessary condition for a molecule to be optically active. Write down the statement here. If a molecule is chiral, if a molecule is chiral, then it is optically active. It is optically, it is the necessary and it is the necessary and sufficient condition and sufficient condition for a molecule to be optically active, to be optically active. Keep that in mind, okay? If somebody says, what is the condition for a molecule to be optically active? Your answer should be, your answer should be, the molecule should be chiral in nature, okay? The molecule should be in chiral in nature, right? If it is not chiral, then it is optically inactive, okay? Write down next, chirality has nothing to do with, has nothing to do with, one second, Ayutthaya, has nothing to do with chiral center. Chiral center is completely different from this particular term. You're talking about this one, Ayutthaya? This one, how? You place this molecule over it. This overt is coming over here. No, that's fine. But this edge is towards the right of the clody. If you consider this, you have to rotate the molecule out of the plane, then it is possible. No, no, see, it is below the plane, obviously. But suppose, if this two bond we have, below the plane, right? This is below the plane, but the direction is Ulta, no? The direction is Ulta. You can also have two bonds like this below the plane, it is going, right? So you can say this is on the right side. So this is on the left side, both below the plane only. So they won't overlap. No, replace, you cannot replace OH and H. That's what replacement for that only, we rotate the molecule. So if you replace, means you're flipping the molecule, Hariharan. If you replace the position of H and OH, means what? You're flipping the molecule, which is not allowed. You cannot flip it, you can rotate the molecule. That's a different thing. If you're talking about different molecule, that molecule that you're talking about, it is not a chiral molecule. It is a chiral molecule. So if you change the molecule, then obviously we have to see from the beginning, what happens and what not happens. I'm talking about this molecule here. For this molecule, the one that you're talking about, it is not chiral molecule. It is a chiral, that's right. Yeah. Okay. So the important point is what if you, the condition of optically active compound that it should be chiral in nature. What is chiral compound? There is no elements of symmetry. Means we cannot divide the molecule into two equal halves. That is the chiral nature of a compound here. Now write down next chiral center. chiral center, write down it is, it is an sp3 hybridized atom. It is an sp3 hybridized atom having four different atoms or groups, four different atoms or groups attached to it. Okay. Condition is the atom must be sp3 hybridized and all the four atoms or groups must be different. Okay. For example, you see, I have this molecule, C, C. We have here NH2, we have here H, we have here OH, here CL, we have here H, we have here H, we have here OH, we have here H. Randomly I have drawn one molecule. Okay. If somebody asks you, if somebody asks you, how many chiral centers we have in this? Could you answer this? How many chiral centers? Okay. How many chiral centers we have in this? Two chiral centers. Because if you look at this carbon atom, fine, it is sp3 hybridized. Okay. But it has two same atom, hydrogen attached to it. It is not a chiral center. If you talk about this, again, it is sp3 hybridized. Okay. And if you look at the atoms or groups attached to it, one is hydrogen, other one is chlorine. Third group is this group, the entire group. Okay. It's third. And this is the fourth one. Obviously you see this group and this group are different because one contain OH, other one contain NH2. Correct? This, all the four atoms or groups are different for this chlorine. So it is a chiral center and chiral center we represent like this is a star mark on this carbon atom. Similarly, if we talk about this carbon, it is also a chiral center, isn't it? So we have two chiral center present in it. Yes, no doubt. If carbon is the chiral center, then we call it as chiral carbon. chiral carbon. Okay. Okay. What is important here? What is important? Both hands are now. Thing is, if the molecule contains, write down this, if the molecule contains, if the molecule contains only one chiral center, only one chiral center, then the molecule is optically active, is optically active. Okay. What I said, molecule contains only one chiral center. Okay. More than one, we do not know. Right. It is the condition for only one chiral center. If you have more than one chiral center, then it may or may not be optically active, we are not sure with it. Okay. Again, the important point here is what? Presence or absence of chiral center, absence of chiral center is not any criteria for a molecule to be optically active. What is the criteria for optically active molecule? What is the criteria? Yes. The molecule must be chiral. Very good. The answer is it is a chiral molecule. It should be a chiral molecule. It is not like there should be one chiral center present. No. The condition for optical activity is the molecule must be chiral in nature, right? And chirality has nothing to do with the presence or absence of chiral center in the molecule, right? But what we have observed, we have observed that if the molecule contains only one chiral center, it is optically active. And with this data, with this information, you can solve almost 90% of the question, okay? Whenever you have to find out whether the molecule is optically active or not, you check for chiral center first. If it has only one chiral center, it is optically active. If it is more than one, then we cannot see. Then what we need to do? We need to check the chirality of the molecule. And what is chirality? Chirality, we need to check whether we can divide the molecule into two equal halves or not. Do we have any plane of symmetry present or not in the molecule? Is it clear? Right? So keep that in mind. chiral center is different, chirality is different, okay? chiral molecule is a molecule which cannot be divided into two equal halves. If there is any kind of symmetry, then the molecule is said to be a chiral molecule, A-C-H-I-R-A-L. If there is elements of symmetry present, the molecule is said to be a chiral, write it out. Next slide down. There's one more term here. Write down stereo center. Write down any atom at which, at which an interchange of, an interchange of groups produces a stereoisomer. A stereoisomer is called, is called stereocenter or stereogenic center. Stereogenic center. Both are same thing. Stereocenter or stereogenic center. Like for example, you see C double bond C, C-L-H-H. One second guys, I'm coming. Yeah. So this molecule you see, we have hydrogen here. It is cis or trans, this one. This is cis or trans. This is cis, right? Because identical molecules, atom, chlorine on the same side, hydrogen on the same side, it is cis molecule, right? Okay. So this is cis. And if you interchange this chlorine and this hydrogen, then what happens you see? If you interchange, then we get C double bond C, H, C-L, C-L-H. Opposite side, right? It is trans. So these two are what? These two are stereoisomers. Hence this atom you see here, this atom is a stereocenter because across this, the atom has been exchanged, stereocenter, right? So how many stereocenter we have in this? One and two. We have two stereocenter in this one. Important point here you have to keep in mind, the stereocenter could be, stereocenter may be sp2 or sp3 hybridized, okay? Remember, carol center is always sp3, but a stereocenter can be sp2 or sp3 as well, right? So we can say all carol centers are stereocenter, but converse is not true, right now. All carol centers are stereocenter, but the converse is not true. Now you see the another one. These three terms you must take care of, okay? Representation of molecule, we have next fissure projection, fissure projection. What is fissure projection? You see, we have this molecule. The name is butane 2,3 diol. We can draw the structure of it. Butane is three carbon, sorry, four carbon. And with second and third carbon, we have OH present, right? OH, OH, here we have H, here we have H, and all these we have hydrogen, CH3 and CS. No doubt in this, butane 2,3 diol is this. Now what I'm telling you, you just rotate this molecule. You just rotate this molecule 180 degree, clockwise or anticlockwise, anyone? 180 degree, clockwise or anticlockwise, okay? 180 degree, anticlockwise, I'm assuming here, correct? What happens? You'll get this, isn't it? This CS3 will be here on the top. This CS3 will be on the bottom. OH, OH, H or H. Any doubt in this? This structure is fissure projection, this one. It looks very simple, but it is not actually. The actual molecule, yeah, that's what I'm coming to that point, Arita, once again. Like I said, it looks simple, but it is not actually, okay? It looks like planar, but it is not in fact, okay? Now what is this? All the horizontal lines are coming out of the plane towards the observer. What I said, all horizontal lines coming out of the plane towards the observer. So if you want to draw the actual structure of this molecule, right? The actual structure, how it looks like, this would be this, all the horizontal bond I'll draw this way, just a second. We have this carbon and carbon bond, vertical line, middle one, right? OH is on the right, but it is coming out of the plane like this. OH, H is also on the left but out of the plane like this. OH out of the plane and H out of the plane. All the vertical lines, this well of bond, it is going into the plane, CH3 and CH3. So this is what the Fisher projection is. It looks like planar, but it is not, okay? So when you see this structure, you should imagine this one, then only you will understand that how the actual structure looks like. Copy this down, understood? Done, okay, yes, whatever. This is a kind of, you can say, reference we have. We assume like this only. We define Fisher projection like this only, okay? So this is, we can say, we have the representation like this. So you can understand it this way only. This is how it is defined actually. There's no definition. You can say it is the, you know, the reference molecule is this we are taking. We define Fisher like this only, okay? Yeah, fine. Next you see, so we have discussed two different representation of molecule. What are they? Two different representation. That is Fisher and flying wedge, yes? Fisher and flying wedge representation, right? Yes, now remember in the beginning, he said that if it rotates a plane polarized light towards the right, it is D, dextrinated tree. Towards the left, it is levorated tree. D is plus and levo is minus, remember that, okay? We cannot say a given molecule theoretically, we cannot say a given molecule is dextro or levo. And for that what we need to do, we need to perform that experiment in the lab, isn't it, right? But I have told you that we have a theoretical method by which we can define whether, define the behavior of the molecule towards the plane polarized light, clockwise or anti-clockwise. And that we call it as RS configuration. Theoretical way it is, RS configuration. Remember this RS has nothing to do with D or L, correct? R can be D, R can be L also. Similarly for S also, right? This R and S has no relation with dextro or levo. Don't get confused here, right? So this is the theoretical method. Practically what happens, we do not know, okay? For that we have to do that experiment. What is R stands for? R stands for, we have a Latin word called rectus. Not important, just you write it down. Rectus and rectus means right. S stands for, again a Latin word called sinister. And sinister means left, okay? So before going into this, like what is R? What is S? How do we define R or S? Okay, two conditions you just understand here. You can have two different representation of molecule. Either they will give you flying wedge representation or they will give you Fisher projection formula, correct? So if you have a Fisher projection formula, let's say this one. This is a formula we have, for example, okay? We have only one carol center in this. That is also possible, right? In the previous example that we took, in that how many carol center we had? Could you check once, all of you? How many carol center we have in the previous one? Yes, pre Fisher projection, previous one. How many carol center? Two, the middle one, the carbon. Both middle and the carbon, which has OH attached, right? Yes, two carol center. Here we have only one carol center, what a big deal, right? Okay, so a cross, that is it, okay? This is Fisher, okay? So it has supposed some group attach A, B, C, D. Randomly I'm giving you this A, B, C, D, right? Or we have flying wedge representation. For example, this A, B, C, D, correct? A, B, C, D, P, P and Q, because C is already there, okay? Now to assign this configuration R and S, what you have to do, you have to assign the configuration on the atoms or groups attached to the carol carbon, according to CIP rule, correct? We have to assign the priority. Just go back and check CIP rule quickly, all of you. According to CIP rule, we'll assign the priority, okay? So condition is what? Condition is, if you have a Fisher prediction, then we must have least priority group on the bottom of the vertical line, this one, because one, two, three, four, fourth one is the least priority. So fourth must be here. This could be anything, one, two, three, it could be anything, that's not a condition. But condition is this, fourth could be here. Similarly, if you have flying wedge representation, so fourth must be in the bottom of the vertical line. So it is the least priority group, least priority. Is it clear? This is the condition we have. Now you'll see how to assign the configuration here, okay? Same example I'm taking, carbon with four different groups attached. So we have, suppose here's, I'll write down randomly PQRS with P, Q, R and S. So we'll assign the priority. Randomly I'm assuming first priority is this, second priority is this, third priority is this, and we know this must be fourth. If it is not, then what to do? We'll discuss that later. But here we are assuming in, like into the plane, we have the least priority four, okay? We assign the priority, then what we'll do, you see? We'll go from the first priority group to the third priority group. First to third we have to go. There are two different ways. Ek toh, you can go directly like this. One say four, four say three. Other one is this. One to two first and two to three. So which path we have to follow? We have to follow the second one. Means first of all, we have to go from first to second and then second to third. First to second and then second to third will go. Not like this. All these are methods, okay? All these are methods technique given to assign the configuration, okay? So if this is, if this, you know, path one to two, two to three, if this path is clockwise, if you're getting it as clockwise, then it is our configuration. If this path is anti-clockwise, anti-clockwise, then it is S configuration. This is how we assign the configuration in molecules. Understood? Any doubt? Whatever it is. In this case, it is anti-clockwise, but in other molecules, it can be clockwise also, no? I'll do it like this. Just one thing you see. Suppose what I did, I put R here and Q here. Now what happened? It's three, it's two. Then what is this? Clockwise? Yes or no? So it can be anything in the molecule. Clockwise, R, anti-clockwise, S. So I change the numbers. Clear? No doubt? Okay. So this is what we need to do to find out the configuration of a molecule. We'll see some examples now, you see. Assign R and S configuration. Carbon, we have CH3, we have VR, carbon. Keep some space in between, okay? We have to discuss a lot of things over here. CH3, CL, BR, CH3, CL, CS3, BRH, okay? C, BR, CH3, CL, H, CS3, BRH, CL. Find out the configuration of all the compounds. It is one, it is two, three, four, R and S configuration. Apply CIP rule, assign the configuration and then you follow the path, one to two, two to three. Clockwise, R, anti-clockwise, S. Done guys? Okay. Okay. So this is what we need to do. Two, two, three. Clockwise, R, anti-clockwise, S. Done guys? Should I do the first one? Okay. Why neither, Aritya? Could you think about it, the last one? Okay. Last one, optically active or not? Last one, optically active or not? Optically active, okay. Fifth one is, when it is optically active, it must be either R or S, isn't it, Aritya? Yes? So we can assign the configuration, right? Okay, one more thing. How do you know it is optically active? One canal center, right? Fine. If it has only one canal center, it is optically active. You can also think of, there is no plane of symmetry. Whatever you want, you can think. You won't get any plane of symmetry over here, right? So in this molecule, the first one, I'll talk about it, how to assign configuration in this one. Okay. We'll start from the first one. Tell me the priority order. Priority order of CS3, BR, CL, H. Can we say bromine has the highest priority because maximum atomic number? Yes? We compare the atomic number. Yes or no? Guys, could you respond, all of you? Right? So one, then two, then four, and then three, isn't it? So we have this condition that least priority group must be in the, into the plane, right? So this is fine. We don't have to do anything. So we have to go from one to two, and then two to three. One to two, it is this, then two to three, it is this. So it is anticlockwise. Anticlockwise means the configuration is, now in this one, we have one, then we have two, then we have three, and then we have four. So in this one, we have one to two, and then two to three. Clockwise are one, two, three, four. One, two, three, and four. This one is also one, two, three, and four. Okay? Now what we have to do next, you see, one, two, two, and two to three. One, two, two, and two to three. This one you let it be now. It is anticlockwise, S, this is clockwise, R. Any doubt in this four? Any doubt in this? All of you, no doubt. Abduhu, the answer for this question is there in this, for example, if you observe this four, you can answer this one as well. What I'll tell you. Achha, you see, you have this compound, just to ignore all these, you know, configuration and priorities, everything, you know, just you look at this compound in which carbon atom is attached with methyl bromine, chlorine, and hydrogen. And this one also you see. Could you tell me, how do you get this molecule, the second one, from the first one? How do you get, tell me, how do you get the second one from the first one? If you observe, if you swap chlorine and bromine, isn't it? Swap CL, NBR. How do you get this from the second one? If you swap CL and CS3, isn't it? CL and CS3. How do you get this one? By swapping, what? Tell me, swap what? CL and BR. Achha, okay. See, observe this. When only one pair is getting swapped, like only CL and BR, only one pair, so one time swap we have. Here also we have one time swap, only one time swap we have, correct? So if you have only one time swap, then the configuration changes from S to R, R to S, S to R. Can we say that? Yes. Azul, one time you are not very much sure, right? Clearly you can see, only one pair exchange we have, only one time exchange we have, S converts into R, R converts into S, S converts into R. Now, if you think of this one, from this to this, how do you get, how many swaps we have, just you tell me. Don't draw this, just you tell me. How many swaps we have from first to fourth? How many swap? One swap here, then two, and then three, isn't it? Three swap? What, would you answer? Yes, sir, no, tell me guys. Just one, that way also you can think, ha, fine. One swap also, just you let it be. Direct, I'm not going. If you consider from here to here in this path, how many swap we have? This also you can do, that's right. This path you tell me, how many swap we have? Three, correct. So if you have one swap, then configuration changes. If you have three swap, then configuration changes S to R. Then if you think of five swap, then also S becomes R, R becomes S. So what we can conclude, we can conclude here, whenever you have odd number of swap, one, three, five, seven, then the configuration changes from S to R or R to S. Is it clear to all of you now? Yes, odd number of swap, configuration will change from S to R or R to S, whether it is one, three, five, seven, whatever. Right, that's why you see, only one swap, S becomes R. From here to here, one swap, R becomes S. One swap, S becomes R. Here to here you see, if you look at direct from this one to this one, only one swap we have, S becomes R. If you consider this path, one, two, three. Three swap, S becomes R. So we can conclude from this, that whenever we have odd number of swap, configuration changes from S to R or R to S. You can write down this particular point in your own word. Write it down first. One second, one second, we'll do, we'll do. First you write it down, whatever I said. Okay, odd number of swap, configuration changes. Now I want you to consider first and third molecule. How do you get third from the first? How many swap we have? One to third, if you are going, how many swap we have? Look at it, it's simple, right? From here to here, one swap. From here to here, one swap. So two swap we have. Two swap, right? All of you guys, respond. Two swap, yes, can we say two swap? Yes, so when we have even number of swap, you see, S is S only, there's no change. Any doubt? If you want to see one more, look at these two, look at these two. How many swap? One and two, R is R, no change. Tell me, any doubt here? Till here, any doubt? Even number of swap, configuration changes. Sorry, configuration does not change. Odd number of swap, configuration changes, correct? Now this point only, this logic only you apply to find out the configuration of this molecule, okay? What we'll do here, listen to me very carefully, what we'll do here, since we have the condition that least priority must be into the plane. So what I will do, whatever the atom is present into the plane, right? This is CL is present into the plane. So I will replace the least priority one, exchange the least priority one with the atom or group which is going into the plane. This is for example, H is the least priority. So I'll replace this H and CL, what will we get? We got H here, we got CL here, and this one would be as it is, this one would be as it is. Give an example that we have done already, this one, correct? So since we replace this, we exchange this, now we can find out the configuration of this one. What is the configuration of this one we have already done? The configuration of this one is S. So if this one is S, so from here to here you need to go. How many exchange we have? How many exchange we have? One, odd, odd exchange, so S becomes what? Tell me, think about it. Let me know if you have any doubt, clear? Like this, we can find out the configuration R and S here. You should be very clear with the priority, like what priority we need to assign when, what is the condition we need to apply? Once you have assigned the priority, then it is nothing, clear? Can we move on now? Similarly, you can also think of in Fisher prediction, very simple example I'm just giving you. For Fisher, you see, like this the molecule would be, right, this is obviously carbon, suppose this is OH, this is CS3, this is NH2, and this is H. So least priority must be here on the bottom of the vertical line, okay? That is the condition we have. Tell me the configuration of this molecule. Yes, yes, yes, configuration means R and S only. What is the priority of OH here? Priority of OH? I know you are having difficulty in this priority thing only. How many of you are comfortable with assigning priority? Those who are not comfortable, just go through the CIP rule we have done in geometrical isomerism, correct? What we do in this one? What we do in this one? If you have any group, we'll compare the atomic number of the first atom, like O, C, N. If it is atom, no problem, atomic number we'll compare. More atomic number, more priority. When we have isotope, then we compare the mass number, okay? So oxygen has maximum atomic number one, then we have nitrogen, then we have carbon, and then we have hydrogen. If hydrogen is not there, we have to exchange, swap the position, right? So we have to go like this. It is clockwise or anticlockwise. You see, we are getting it as clockwise. Okay, now we'll move on. Next term we have enantiomerism, right down the term, right on the compounds which are, the compounds which are, the compounds which are non superimposable, mirror image of each other, are called enantiomers. Compounds which are non superimposable, mirror image of each other, are called enantiomers. And this phenomenon is enantiomerism. And this phenomenon is enantiomerism. Yeah, the compounds which are non superimposable, are non superimposable, mirror image of each other, non superimposable, mirror image of each other, are called enantiomers. And this phenomenon is enantiomerism. Simple example, you can take a trans isomer, any trans isomers, you see this. Suppose we have carbon, carbon double bond, and it is a trans molecules, both chlorine. It's present opposite to each other. H and H, okay? If you take mirror image, because the definition is what? Non superimposable mirror image of each other. So we have to find out the mirror image first, and then we can check whether it is superimposable or not. It's not like you have to do all these things every time when you solve the question. Once you practice a bit, you can understand that whether it is enantiomers or not, okay? So I'm taking mirror image of this. What will it be? C double bond C as it is written here. This hydrogen is here, so hydrogen, chlorine. This chlorine is close to the mirror, at this edge. Tell me, these two molecules are superimposable. Check once. Are these molecules superimposable? No, right? Even if you can think of by rotating the molecule also. One is so that right you pick this, this molecule you pick and place over it, right? You see this edge will come over this chlorine, right? Not superimposable. If you think of rotating this, then also it's not possible. You can think of, imagine this, you'll get it, right? So these are what? These are non superimposable mirror image. Hence, these are the pair of enantiomers, enantiomers. Next, there's a compound called glyceraldehyde. Have you heard this name? Glyceraldehyde. No? Yes. In this molecule, what happens? We have one carbon atom attached with C-H-O, primary functional group is C-H-O. And then we have OH, H attached to it. And then we have C-H-O-H. This molecule is glyceraldehyde. Do we have carol carbon in this? Do we have carol carbon in this? What is carol carbon? What is carol carbon, guys? Yes. You can think of this carbon here. SP3 hybridized, four different group attached to it, right? So this carbon is what? This carbon is carol carbon. Can we say that? Without any doubt. Centrival of. This molecule is carol or not? Is this molecule carol? Yes. And why it is a carol molecule? Because we do not have any symmetry in this, okay? It is carol not because it has a carol center. It is carol because there is no symmetry in this. Since it is carol or it has only one carol center, it is optically active, right? Optically active. If you take the mirror image of this, right? Then the mirror image is also non-superimposable because the carol center. So mirror image and this molecule is what? Is again, enantiomers of each other. See this. This is the mirror image we get. These two are non-superimposable. No matter whatever you do, or how you rotate the molecule, you get the same thing. Don't think about that we'll rotate it. Obviously flipping is not allowed. You cannot flip the molecule, right? And hence these two compounds are enantiomers of each other. Both are optically active, right? So one will be dextro, other one will be levo. We don't know which one is dextro, which one is levo exactly. But if this one is dextro, this one is levo, or this one is dextro, this one is levo, right? Right down all enantiomers are optically active. All enantiomers are optically active properties, you write down. Yes, Pooja you were asking SN2 reaction, right? In the doubt. Yes, so after finishing this, optical will do that. Once we start hydrocarbon, we'll do that, okay? Yeah, so all enantiomers are optically active, okay? Now very important point, write down note this point is, if you take equimolar mixture of both compounds, write down like this. If you take equimolar mixture of, if you take equimolar mixture of enantiomers, then the mixture will be optically inactive. Then the mixture will be optically inactive. And this is called resmic mixture, R-A-C-E. Means what I'm telling you here, suppose if you take N mole of this compound, equimolar mixture and N mole of this compound, if you mix the two, then this mixture is called resmic mixture, R-A-C-E, M-I-C, resmic mixture, the term we use, okay? This compound is optically active, this compound is optically active, but this one, resmic mixture is optically inactive. Previous one, the previous one you're talking about, this one, yes enantiomers are optically active, one compound is optically active, mixture is not, resmic mixture is not optically inactive. Next slide down, the next term we have in this is di-stereomers, di-stereomers, write down stereoisomers, stereoisomers which are not mirrored image of each other, stereoisomers which are not mirrored image of each other are called di-stereomers, okay? Stereoisomers which are not mirrored image of each other are called di-stereomers, right? Next point to write down, see cis and trans cannot be the mirrored image of each other, right? If you take one cis molecule and one trans molecule, they cannot be the mirrored image of each other, right? So di-stereomers are what are all geometrical, right down, all geometrical isomers, isomers are the example of di-stereomers, all geometrical isomers. Next slide down, di-stereomers in a pair, these are properties you should know, di-stereomers in a pair can be of any combination, can be of any combination, means both di-stereomers can be optically active, both can be optically inactive, and one optically active, and one optically inactive, any possibility we have. Next slide down, di-stereomers, different physical properties, different physical properties like, they have different boiling point, different melting point, different refractive index, that is mu, all these properties are different, and since their physical properties are different, they can be separated by any physical methods, like fractional distillation. Hence, can be separated by any physical methods, like fractional distillation. This is not in our syllabus, but sometimes they ask this question, okay? Chromatographic crystallization also possible, chromatographic crystallization. Just these properties you have to memorize, okay? Nothing else, okay? Last one in this, right down, calculation of stereo isomers. We have formula like we had in geometrical isomers, use that formula, you'll get the answer. First one in case of unsymmetrical molecule, there are two parts we have in this, number of optically active form, number of optically active form, we call it as A, and number of optically inactive form, call it as M. So if in case of unsymmetrical molecules, if N is the number of asymmetric carbon atom, asymmetric carbon atom, means stereo center type, you can say N is the stereo center, when the value of A would be two to the power N, two to the power N, and there is no, there is no optically inactive form in this. So totally stereo isomers would be, two isomers, stereo isomers, A plus M, and that is two to the power N, correct? Look at this example, CH3, COH, H, COOH. This carbon you see, it is the asymmetric carbon, four atoms are different, and there is, yeah, I'll go back once again. Okay, so N value is one here, right, total optical isomers would be, A plus M, M is zero, A value is two to the power N, answer is two. Second case, in case of symmetrical molecules, first case in this, when N is even, then the value of A would be, two to the power N minus one, number of optically active form, optically inactive form would be two to the power N by two minus one, total optical isomers, A plus M, two to the power N minus one, two to the power N by two minus one. This is the answer we have. Okay, look at this question on this. This carbon is asymmetric carbon, this carbon is asymmetric carbon. N value is two, okay? Total number of isomers, A value would be, two to the power N minus one, and M value would be, two to the power N by two minus one. This value is one, and this value is two, total answer would be three. You can see this, you can find a symmetry over here, same thing on both sides. Optically inactive forms, we have only one kind of compound in which we have a plane of symmetry. Meso compounds, we call it as, I'll discuss that a bit later, I'll discuss that. Meso compounds, optically inactive form, we call it as meso compounds. The last one we have, when N is odd, number of optically active forms, A equals to two to the power N minus one, minus two to the power N minus one by two. And inactive form is two to the power N minus one by two. This we also call it as meso form. What is meso? I'll discuss, M-E-S-O, meso form, optically inactive form. So total isomers would be, both will add A plus M, two to the power N minus one. Okay. Now, one last thing you see here in this chapter, compound containing, like we need to, just one question we'll see, but this only we'll clear the concept. Four compounds I'm drawing over here. Okay. So this is, we have, we have OH CH3 H OH CH3 H. And I'm taking the mirrored image of this one and mirrored image of this one. Here we have CH3 CH3 OH CH OH CH H. OH CH H. Mirrored image of this one we want. OH here. CS3 here. CS3 here. OH here. H. H. Then we have both CS3 this side. Both OH this side. H on the top. H on the bottom. This is one. This is two. This is three. And this is four. You have to tell me how many enantiomeric pair we have in this? How many diastereomeric pair we have in this? See for enantiomers, what we need to find out? Non-superimposable mirrored image, right? So you see both one and two are mirrored image. Three and four are mirrored image. One and two are non-superimposable. Are they non-superimposable? Yes. Can we rotate this and think whether it is superimposable or not? Like this, if you rotate 180 degree and you place over this, is it superimposable then? So enantiomeric pair, one and two. What about three and four? Are they superimposable? Are they superimposable, three and four? Yes. You can easily rotate this 180 degree, 180 degree and you can place over it, right? Hence these are superimposable, not enantiomers. Only one pair we have for enantiomers. Diastereomers, if you think of, diastereomers are simply not mirrored image of each other. So we can think one and three, not mirrored image, one and four, not mirrored image, two and three, not mirrored image, two and four, not mirrored image of each other. Is it right? Achha, if the question is how many resmic mixture we get? Resmic mixture. How many resmic mixture we get? Only one, very good. Only one resmic mixture we get because we have only one pair of enantiomers. So we'll take five moles of this, five moles of this and mix. So one mixture we get, only one resmic mixture. There's a term called mesocompound. Mesocompounds are always optically inactive, right now. Like you see this compound, there is a plane of symmetry over here. This is a plane of symmetry, right? Because of this symmetry only, it is optically inactive and it is mesocompound. Same thing we have here, right? So three and four are mesocompound we have, right now. Three and four. Mesocompound we don't define in a pair. Three is a mesocompound. Fourth is also a mesocompound, right now. Now one very important property here you must understand. See mesocompounds are optically inactive because of this symmetry, isn't it? Yes, tell me. Yes. This symmetry is passed, like symmetry passes through within the molecule. Molecule can be surpassed and half this side, half that side, correct? So it is passing through the molecule, like divides the molecule into two equal halves, right? That's why we say mesocompounds are optically inactive due to internal compensation. Very important term it is must remember, okay? Because the plane of symmetry is passing through within the molecule. Hence it is internal compensation, clear? If you talk about mesocompound, sorry, resmic mixture, resmic mixture, what is it? One and two are forming. So if you take one and you mix with two, right? It is optically inactive. It is also optically active. So what happens in the mixture? A PPL will go like this. This will rotate clockwise. When it comes to this compound, this will rotate anticlockwise. And finally it deviates without, it comes out without any deviation. So here the compensatory effect we have of the optical active nature of this molecule that is not present within the molecule, right? It is because of this optical activity of this molecule is compensated by another molecule. Same thing we can say this way also. The optical activity of this molecule is compensated by this molecule because both are opposite in nature at least towards the PPL if you see. That's why if you talk about resmic mixture, resmic mixture are optically inactive. We know it is optically inactive due to external compensation here. Is it clear? Due to external compensation. Tell me, any doubt? This through term you must remember internal compensation and external compensation. So this is it for optical isomerism. Okay, we are done with it. One more thing is left in this chapter that is conformational isomerism. Confirmational isomerism will do later after finishing hydrocarbon. Yeah, you should get that either. See what happens? It's not like this compound itself is a mesocompound. It is a mesocompound, right? So we have only one mesocompound in this. See, when you have a molecule, a given molecule which shows optical activity with different, different arrangements, you may get mesocompound, you may get optically active compounds. Like you see, you can think of, what is the name of this compound? Ayupake name, can you tell me? Can you tell me the Ayupake name of this compound? Tell me the Ayupake name of this compound, the four compound that I've drawn, all these compounds have the same molecular formula if you see. What is Ayupake name? It is butte 1,3 diol. It's not butte, it's buttein 1,2 diol. Buttein 1,2 diol with E in the last, E will be there. Okay, buttein may E will be there in the last. So it is buttein 1,2 diol, right? So all the compounds 1, 2, 3, 4 are buttein 1,2 diol. Buttein 1,2 diol, obviously it shows optical activity, correct? Ah, yes, sorry, I'm also sorry, 2,3 diol, yes. All these are, if you want to count the number of optical isomers for buttein 2,3 diol, okay? You can count with that formula. And in this, we have few number of mesocompounds and few number of optically active compounds. That is what it is drawn over here. So third and fourth, if you see, it is already a mesocompound. You can use the formula to count this, fine. Okay, so this is it for optical isomers. Okay, we'll start hydrocarbon, we'll finish that first and then we'll do conformation. Till, what did we have class? Any idea? Anyways, no problem, we'll have time. At least we have two, three classes we'll have. We'll get, two classes definitely we'll get, I guess. Anyways, so before starting hydrocarbon, we'll take a break now. Okay, we'll take a break now and then we'll continue with hydrocarbon. So after the break, we'll start with hydrocarbon. After finishing hydrocarbon, we'll do this, conformation isomers also in the last. Okay, take a break now. We'll resume the session at six, almost six, 15 will resume, okay? 18 minutes break we are taking. Six, 15 will resume, take a break.